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In Kelley-Namioka "Linear Topological Spaces",Page 31, it gives two equivalent definitions of projective topology(i.e.,weak topology). I don't know why they are equivalent.

The point is I don't kown why the following statement is equivalent to the definition of the weak topology:

"The projective topology(A.K.A. the weak opology) can be described alternatively by specifying that a subset $U$ of $X$ is open relative to te projective topolgy if and only if $U$ is the inverse under $P$ of an open subset of $\prod_{f\in F}{Y_f}$, where $P$ is the map which sends a point $x$ of $X$ into the point with $f$-th coordinate $f(x)$(that is, $P(x)(f)=f(x)$)"

I hope someone can tell me how to prove it.

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To clarify the question to those who don't have access to Kelley-Namioka: This is about the initial topology of a family $F$ of maps $f \colon X \to Y_f$ from a set $X$ to topological spaces $Y_f$. The question is why the initial topology (the weakest topology making all $f$ continuous) can be equivalently described as the weakest topology on $X$ making the evaluation map $P \colon X \to \prod_{f\in F} Y_f$ continuous. – t.b. Sep 15 '12 at 2:10
No,$P$ is not a evaluation map, it maybe a representation of x. $P$ should be in $X^{**}$. – Chao Sep 15 '12 at 2:18
We are only talking about topological spaces, there are no dual spaces here, so $X^{\ast\ast}$ makes no sense. The map $P \colon X \to \prod_{f \in F} Y_f$ is called the evaluation map on Wikipedia because it evaluates the functions $f$ simultaneously. $P \colon x \mapsto (f(x))_{f \in F} \in \prod_{f \in F} Y_f$. – t.b. Sep 15 '12 at 2:21
If you think my comments are incorrect and that I misunderstood your question then please add both definitions to your question in order to make this question self-contained. – t.b. Sep 15 '12 at 2:31
Very good :) Welcome to the site! – t.b. Sep 15 '12 at 8:38
up vote 2 down vote accepted

Suppose that $F$ is a family of maps from a set $X$ to spaces $Y_f$ indexed by $F$. Let $Y=\prod_{f\in F}Y_f$. Let $$P:X\to Y:x\mapsto\langle f(x):f\in F\rangle$$ be the evaluation map. One way to show that the weakest topology making each $f\in F$ continuous is the same as the weakest topology making $P$ continuous is to show that a topology $\tau$ on $X$ makes $P$ continuous iff it makes each $f\in F$ continuous.

Suppose that $\tau$ is a topology on $X$ making each of the maps $f$ continuous, so that for each $f\in F$ and each open set $U\subseteq Y_f$, $f^{-1}[U]\in\tau$. Then $U$ is the union of basic open sets of the form

$$B=\{y\in Y:\forall f\in F_0(y_f\in U_f)\}\;,\tag{1}$$

where $F_0$ is a finite subset of $F$, and $U_f$ is an open set in $Y_f$ for each $f\in F_0$. Thus, to show that $P^{-1}[U]\in\tau$, it suffices to show that $P^{-1}[B]\in\tau$ for each $B$ of the form $(1)$. But

$$\begin{align*}x\in P^{-1}[B]&\text{ iff }P(x)\in B\\ &\text{ iff }P(x)_f\in U_f\text{ for each }f\in F_0\\ &\text{ iff }f(x)\in U_f\text{ for each }f\in F_0\\ &\text{ iff }x\in\bigcap_{f\in F_0}f^{-1}[U_f]\;, \end{align*}$$

so $$P^{-1}[B]=\bigcap_{f\in F_0}f^{-1}[U_f]\;.$$

By hypothesis each of the sets $f^{-1}[U_f]\in\tau$, so $P^{-1}[B]\in\tau$ as well, and so is any union of such sets.

Conversely, suppose that $\tau$ makes $P$ continuous. Then $P^{-1}[B]\in\tau$ for every basic open set in $Y$ of the form $(1)$. In particular, for each $f\in F$ and any open set $U_f\subseteq Y_f$, $B=\{y\in Y:y_f\in U_i\}$ is open in $Y$, and therefore $P^{-1}[B]\in\tau$. But

$$\begin{align*} P^{-1}[B]&=\{x\in X:P(x)\in B\}\\ &=\{x\in X:P(x)_f\in U_f\}\\ &=\{x\in X:f(x)\in U_f\}\\ &=f^{-1}[U_f]\;, \end{align*}$$

so $f^{-1}[U_f]=P^{-1}[B]\in\tau$.

This shows that the topologies making $P$ continuous are exactly the same as the topologies making every map in $F$ continuous, and it follows immediately that the weakest topology making $P$ continuous is the weakest topology making every map in $F$ continuous.

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Thanks for your proof, i've got it. – Chao Sep 15 '12 at 6:59

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